Question: $\overline{AC}$ is $10$ units long $\overline{BC}$ is $5$ units long $\overline{AB}$ is $5\sqrt{5}$ units long What is $\cos(\angle ABC)$ ? $A$ $C$ $B$ $10$ $5$ $5\sqrt{5}$
Solution: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{BC} = 5$ hypotenuse $= \overline{AB} = 5\sqrt{5}$ $\cos(\angle ABC )=\frac{5}{5\sqrt{5}}$ $=\dfrac{ \sqrt{5}}{5}$